Question

A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm², respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

Answer 1

Given, the length of the rod is 1.05 m, the cross sectional area of the steel wire A and the aluminum wire B is 1.0  mm 2 and 2.0  mm 2 respectively.

The given description is shown in the figure below,

(a)

Let F 1 and F 2 be the tension in the wire A and B respectively. Let A 1 and A 2 be the cross sectional area of the wire A and B respectively.

The stress in the wire is given by the equation,

stress= F A

The stress in both the wires is same. Therefore,

F 1 A 1 = F 2 A 2

Substituting the values in the above equation, we get:

F 1 1.0  mm 2 = F 2 2.0  mm 2 F 1 F 2 = 1 2

Taking the moment of tension in both the wires about the point at which the mass is suspended, we get:

F 1 x− F 2 ( 1.05 m−x )=0 ( 1.05 m−x ) x = F 1 F 2

Substituting the values in the above equation, we get:

( 1.05 m−x ) x = 1 2 x=0.7 m

Hence, the mass m should be suspended along the rod at a point 0.7 m from the wire A.

(b)

The Young’s modulus of the steel wire and the aluminum wire is 2× 10 11  N/ m 2 and 7× 10 10  N/ m 2 respectively.

Let Y 1 and Y 2 be the strain in the wire A and B respectively.

The strain in wire A is given by the equation,

strain= F 1 A 1 Y 1

The strain in wire B is given by the equation,

strain= F 2 A 2 Y 2

The strain in both the wires is the same. Therefore,

F 1 A 1 Y 1 = F 2 A 2 Y 2

Substituting the values in the above equation, we get:

F 1 1.0  mm 2 ×2× 10 11  N/ m 2 = F 2 2.0  mm 2 ×7× 10 10   N/ m 2 2 F 1 F 2 = 10 7

Taking the moment of tension in both the wires about the point at which the mass is suspended, we get:

F 1 x− F 2 ( 1.05 m−x )=0 ( 1.05 m−x ) x = F 1 F 2

Substituting the values in the above equation, we get:

( 1.05 m−x ) x = 10 7 x=0.43 m

Hence, the mass m should be suspended along the rod at a point 0.43 m from the wire A.